2017CCPC网络选拔赛 1003-Friend-Graph

最新推荐文章于 2020-09-22 19:09:02 发布

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最新推荐文章于 2020-09-22 19:09:02 发布

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分类专栏： ACM算法训练｜蓝桥杯文章标签： 2017CCPC 网络选拔赛 2017CCPC 网络选拔赛1003 F 1003 Friend-Graph CCPC

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本文介绍了一个基于六度分离理论的算法，用于评估团队是否为好团队。通过输入成员间的朋友关系，判断是否存在小团体现象。适用于不超过3000人的团队评估。

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Friend-Graph

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6514 Accepted Submission(s): 1610

Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.

Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.(

n≤3000)

Then there are n-1 rows. The

ith row should contain n-i numbers, in which number

aij represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.

Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.

Sample Input

Sample Output

Great Team!

用到的是六度分离理论：世界上任意六个人中,一定有三个人互相认识,或三个人互相不认识，或者参考 百度百科：六度分离理论

注意，这道题目里面，朋友的朋友不是朋友！！！

代码：

#include <cstdio>
#include <cstring>
#define maxn 6  //只考虑 n= 3,4,5 的情况即可，暴力
using namespace std;

int a[maxn][maxn];  //邻接矩阵 存储无向图
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        int flag=0;
        scanf("%d",&n);
        if(n>=6)
            flag=1;
        if(n==1 || n==2)
            flag=2;

        memset(a,0,sizeof a);
        int value;
        for(int i=1;i<=n-1;i++) {
            for(int j=1;j<=n-i;j++)
            {
                scanf("%d",&value);
                a[i][i+j]=1;
                a[j+i][i]=1;
            }
        }
        if(flag == 2)
            printf("Great Team!\n");
        else if(flag == 1)
            printf("Bad Team!\n");
        else if(flag == 0){
            for(int i=1;i<=n;i++) {
               for(int j=1;j<=n;j++) {
                  for(int k=1;k<=n;k++)
                  {
                      if(i==k || i==j || j==k)
                        continue;
                      //下面是 符合题目中说的， Bad 的两种情况
                      if(a[i][j]==1&&a[i][k]==1&&a[j][k]==1)  {  flag=1; break;  }
                      if(a[i][j]==0&&a[i][k]==0&&a[j][k]==0)  {  flag=1; break;  }
                  }
                  if(flag) break;
                }
                if(flag) break;
            }
            if(flag == 1)  printf("Great Team!\n");
            else    printf("Bad Team!\n");
        }
    }
    return 0;
}